Precalculus Examples

$x^{3} - x^{2} - 4 x + 4 < 0$

Step 1

Convert the inequality to an equation.

$x^{3} - x^{2} - 4 x + 4 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor out the greatest common factor from each group.

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Step 2.1.1

Group the first two terms and the last two terms.

$(x^{3} - x^{2}) - 4 x + 4 = 0$

Step 2.1.2

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x - 1) - 4 (x - 1) = 0$

Step 2.2

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$(x - 1) (x^{2} - 4) = 0$

Step 2.3

Rewrite $4$ as $2^{2}$ .

$(x - 1) (x^{2} - 2^{2}) = 0$

Step 2.4

Factor.

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Step 2.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$(x - 1) ((x + 2) (x - 2)) = 0$

Step 2.4.2

Remove unnecessary parentheses.

$(x - 1) (x + 2) (x - 2) = 0$

Step 3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 1 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 4

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 6.2

Add $2$ to both sides of the equation.

$x = 2$

Step 7

The final solution is all the values that make $(x - 1) (x + 2) (x - 2) = 0$ true.

$x = 1, - 2, 2$

Step 8

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$1 < x < 2$

$x > 2$

Step 9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 9.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 9.1.2

Replace $x$ with $- 4$ in the original inequality.

${(- 4)}^{3} - {(- 4)}^{2} - 4 \cdot - 4 + 4 < 0$

Step 9.1.3

The left side $- 60$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $- 2 < x < 1$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 2 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

Replace $x$ with $0$ in the original inequality.

${(0)}^{3} - {(0)}^{2} - 4 \cdot 0 + 4 < 0$

Step 9.2.3

The left side $4$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.3

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Step 9.3.2

Replace $x$ with $1.5$ in the original inequality.

${(1.5)}^{3} - {(1.5)}^{2} - 4 \cdot 1.5 + 4 < 0$

Step 9.3.3

The left side $- 0.875$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 9.4

Test a value on the interval $x > 2$ to see if it makes the inequality true.

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Step 9.4.1

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Step 9.4.2

Replace $x$ with $4$ in the original inequality.

${(4)}^{3} - {(4)}^{2} - 4 \cdot 4 + 4 < 0$

Step 9.4.3

The left side $36$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 9.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 2$ True

$x > 2$ False

$x < - 2$ True

$- 2 < x < 1$ False

$1 < x < 2$ True

$x > 2$ False

Step 10

The solution consists of all of the true intervals.

$x < - 2$ or $1 < x < 2$

Step 11

Convert the inequality to interval notation.

$(- \infty, - 2) \cup (1, 2)$

Step 12